uncookedfalcon
Reputation
3,224
Next privilege 5,000 Rep.
Approve tag wiki edits
 Jan27 comment How to prove that volume forms agree on $U_\alpha \cap U_\beta$? Hi self-learner, you are welcome. Please feel free to ignore'top form', I just meant that if we on an $n$ dimensional manifold, all $n+1, n+2, ...$ differential forms are 0, so an expression $f dx_1 \wedge \ldots dx_n$ is top' in the sense it has the maximal amount of wedges. By $d \phi$ I meant the Jacobian matrix of $\phi$, so yes $d\phi$ converts expressions of the form $a_i \frac{\partial}{\partial y_i }$ into $b_j \frac{\partial}{\partial x_j}$, i.e. is the infinitesimal change of coordinates. Jan26 revised How to prove that volume forms agree on $U_\alpha \cap U_\beta$? [Edit removed during grace period] Jan26 answered How to prove that volume forms agree on $U_\alpha \cap U_\beta$? Jan10 awarded Yearling Dec11 answered Intersection between a closed set and $y=x$ on $[0,1]$ Dec5 comment construct non-commutative group with prescribed center Why not take the product with $PGL(2)$? Dec5 answered The set of all fixed points of a continuous function $f:[0,1] \to [0,1]$ , satisfying $f \circ f=f$ , is a non-empty interval? Dec5 comment Probability of hitting a point in a square, maximum of coordinates Draw some level sets for the function $\max(x,y)$ on $[0,1]$ - you'll get a bunch of $L$'s. Then it will be clear how to calculate the area. Dec5 comment Flatness on the affine line for a coherent sheaf @user54369 Nothing to apologize for, but still not true: take something like the trivial line bundle and throw in some junk at a few points (so $M = \mathbb{C}[t] \oplus \mathbb{C}[t]/t$, say) - then at every point except the origin, $M$ looks nice and free, but at the origin it's all messed up. Nov2 comment Matrix of Shift Transform on arbitrary basis You are almost there - you have every right to think of $v_1, \ldots, v_n$ as the standard' basis, so your matrix looks like the identity matrix shifted down one row in all but the last column, where it reads as you wrote. Oct23 comment set of all regular values @SarahT. you are absolutely right, I was being careless. Good catch! Oct22 comment set of all regular values As before it suffices to check on a cover: now take a cover $U_i$ for $N$ with each $U_i \simeq \mathbb{R}^n$, and take a cover $V_{ij}$ for $f^{-1}(U_i)$ with each $V_{ij} \simeq \mathbb{R}^m$, with this it suffices to check that for a map $\mathbb{R}^m \rightarrow \mathbb{R}^n$, the noncritical locus is open. At a point in it $x$, the Jacobian matrix $df$ has full rank - it is a linear algebra exercise that this means some $\dim N$ minor of it has full rank - since this minor's determinant is continuous as a function on $\mathbb{R}^m$, it is still nonzero nearby, so still $df$ is full rank. Oct22 comment set of all regular values Yes, you're welcome. Now use that at $x$, some $\dim N$ minor of the Jacobian has nonvanishing determinant (this works for $\geqslant$ as well). Oct22 comment set of all regular values Okay: you can have an open cover of $M$ by copies $U_i$ of $\mathbb{R}^n$, and you can check that a set $W$ is open iff $W \cap U_i$ for all $i$ - this is just point set topology. With that, you're reduced to answering the question for $\mathbb{R}^n$ - you want say that if $x$ is not a critical value, nearby guys are not either. What does it mean to be critical? All your partials vanish. So at $x$, some $\frac{\partial f}{\partial x_i} \neq 0$, by continuity this is true nearby. Oct22 answered set of all regular values Oct22 answered Confused about short exact sequence involving $\mathcal{O}_{\mathbb{P}^n}$ and $\mathcal{O}_Z$ for $\pi:Z\rightarrow \mathbb{P}^n$ closed embedding Sep24 awarded Autobiographer Aug24 answered Definition of a coordinate vector bundle Aug24 comment Definition of a coordinate vector bundle By 1., certainly each fibre of $B$ can be set-theoretically identified with $\mathbb{R}^n$ over $U_f$, using $\phi_f$. By 2., the induced vector space structure doesn't depend on your choice of $f$. Aug24 comment Convergence in measure implies pointwise convergence? @mathematician That seems true to me - if a sequence $a_i$ doesn't converge to $a$, we can find a subsequence which is always at least $\epsilon$ away from $a$, which is preserved by any subsequence. Perhaps the problem is that you are only given a subsequence which converges almost everywhere, and there are uncountably many such sequences (so the total problem area where we don't converge pointwise need not have measure 0)? (convergence in measure is definitely preserved by taking subsequences)